Derive the volume of a sphere.

[pikapika!]

The forumula for 2¶r can intergrated to make ¶r^2 (at least I think). So can anyone derive the volume of a sphere
4/3¶r^3?

 

[dextercioby]

First thing: the volume of a sphere is...0.

Second: to find the volume of a ball seen as a domain in $\mathbb{R}^{3}$, one could antidifferentiate the expression giving the surface of the ball as a function of its radius.

Daniel.

 

[HallsofIvy]

The forumula for $2\pir$ can intergrated to make $\pir^2$ (at least I think). So can anyone derive the volume of a sphere
$4/3\pir^3$?
It is certainly true that the formula for the circumference of a circle can be integrated to give the formula for area of a circle but that has nothing to do with "deriving" the area formula- unless you mean "differentiating" it!
It is true that the formula for the surface area of a sphere, $4\pir^2$ can be integrated to give the volume of the ball but, again, that is not "deriving the formula".

 

[1800bigk]

to derive the volume of a sphere I think you want to start with the unit ball in 3 space. Then convert to spherical coordinates and integrate the triple intergral. I think I am close, I remember doing this a while back so take it FWIW.

 

[amcavoy]

Why not try a surface of revolution? For example, take a semi-circle and rotate it around the x-axis to find the volume.

$$\text{V}=\pi\int_{a}^{b}f(x)^2\,dx$$

 

[EbolaPox]

I solved this out for practice on my own. I found the volume of a revolved surface. I said $$\text{V}=\pi\int_{r}^{-r} (\sqrt{r^2 - x^2})^2dx$$.
I'm pretty sure that should work thus giving you a simple to evaluate integral of
$$\pi\int_{r}^{-r} \(r^2 -x^2)dx$$.
Should be easy enough. If you have trouble with understanding whereabout the integral came from, realize that it is simply the sum of the volume's of cylinders.
(Note: I hope I typed that latex code right...I've never used it before. Sorry if it comes out wrong.)